Free Math Worksheets, Word Problems and Teaching Resources

Discriminant

noreply@blogger.com (Mr. Math) — Wed, 10 Mar 2010 17:06:48 PST

We have already studied the quadratic formula,

The term which is inside the square root symbol is called the discriminant.

It is used to determine the nature of the roots of a quadratic equation. We can also determine the number of real roots for a quadratic equation with this number. The following table will give us the relation between the discriminant and the nature of the roots.

Discriminant	Nature of roots	Number of real roots
b² - 4ac > 0	Two real roots	2
b² - 4ac = 0	Double root	1
b² - 4ac < 0	No real roots	0

Example 1:

Find the discriminant value for the equation x² + 5x + 6 = 0 and determine the number of real roots.

Solution:

Step 1:

Given equation is x² + 5x + 6 = 0.

It is in the form ax² + bx + c = 0, where a = 1, b = 5, and c = 6.

Step 2:

The discriminant = b² - 4ac

b² - 4ac = (5)² - 4(1)(6) = 25 - 24 = 1 > 0

=> The equation has two real roots.

=> The graph of this equation will touch the x-axis twice.

Step 3:

So, the given equation has two real roots.

Example: 2

Find the discriminant value of x² - 12x + 36 = 0 and determine the number of real roots.

Solution:

Step 1:

Given equation is x² - 12x + 36 = 0.

It is in the form ax² + bx + c = 0 where, a = 1, b = -12, and c = 36.

Step 2:

Discriminant = b² - 4ac = (-12)² - 4(1)(36) = 144 - 144 = 0

=> The equation has a double root.

=> The graph of this equation touches the x-axis in only one point.

Step 3:

So, the given equation has a double root.

Example: 3

Find the discriminant value 2x² + x + 3 = 0 and determine the number of real roots.

Solution:

Step 1:

The given quadratic equation is 2x² + x + 3 = 0.

It is in the form ax² + bx + c = 0 where, a = 2, b = 1, and c = 3.

Step 2:

b² - 4ac = (1)² - 4(2)(3) = 1 - 24 = -23 < 0

=> The equation has no real roots.

=> The graph of this equation does not touch the x-axis.

Step 3:

So, the given quadratic equation has no real roots.

Related Articles:

>> Graphing method to solve quadratic equations

>> Quadratic formula method to solve quadratic equations

>> Perfect Square Method: Detailed Review

>> Solving Quadratic Equations: Perfect Square Method

>> Solving Quadratic Equations: Factoring Method

>> Solving Inequalities: Graphing Method

>> Solving Linear Equations: Graphing Method

>> Linear Equations: Word Problems

Best method to solve quadratic equations

noreply@blogger.com (Mr. Math) — Fri, 19 Feb 2010 15:54:07 PST

So far, we have studied the four methods to solve a quadratic equation. How do we select the best method to solve among these four methods?

The following table will guide us to choose the best method to solve.

Method	When to choose
Graphing	Use if we need only the appropriate solutions. However, we can use it always.
Factoring	Use if we could easily determine the factors or the constant term is zero.
Completing the Square	Use if the equation is of the form x² + bx + c = 0 and b is even. However, we can use it always.
Quadratic Formula	We can use this method to any type of quadratic equation. In some cases, the other methods will be easy to solve.

Example:

The product of two consecutive positive odd integers is 99. Find those two numbers.

Solution:

Step 1:

Given that the product of two consecutive positive odd integers is 99.

We are asked to find the two consecutive positive odd integers.

Let the first integer be x.

Then the consecutive positive odd integer is x + 2.

Given that the product of x and x + 2 = 99

=> x(x + 2) = 99

=> x² + 2x - 99 = 0

So, we arrived at a quadratic equation x² + 2x - 99 = 0.

Step 2:

We can solve the quadratic equation by using factoring.

x² + 11x - 9x - 99 = 0

=> x(x + 11) - 9(x + 11) = 0

=> (x + 11)(x - 9) = 0

=> x + 11 = 0 or x - 9 = 0

=> x = -11 or x = 9

Step 3:

Since we are looking for a positive number, neglect -11.

So, x = 9.

If x = 9, then x + 2 = 9 + 2 = 11.

Step 4:

Hence, the two consecutive positive odd numbers are 9 and 11.

Related Articles:

>> Graphing method to solve quadratic equations

>> Quadratic formula method to solve quadratic equations

>> Perfect Square Method: Detailed Review

>> Solving Quadratic Equations: Perfect Square Method

>> Solving Quadratic Equations: Factoring Method

>> Solving Inequalities: Graphing Method

>> Solving Linear Equations: Graphing Method

>> Linear Equations: Word Problems

Graphing Method: Quadratic Equations

noreply@blogger.com (Mr. Math) — Mon, 15 Feb 2010 15:59:14 PST

In this method, we are going to find the solution or roots of a quadratic equation graphically. The graph of a quadratic equation is always a parabola. The points where the graph of the parabola cuts the x-axis are called the roots of the corresponding quadratic equation.

To graph a quadratic equation, we should graph the related quadratic function. For example, if the quadratic equation is x² + 2x + 5 = 0, then the related quadratic function is f(x) = x² + 2x + 5.

Let us solve few problems on quadratic equations by graphing.

Example 1:

Solve x² + x - 6 = 0 by graphing.

Solution:

Step 1:

Given equation is x² + x - 6 = 0.

Here a = 1, b = 1, and c = -6.

Graph the related function f(x) = x² + x - 6.

The equation of the axis of symmetry is x = - ^b/_2a

x = - ^b/_2a = - ¹/₂₍₁₎ = - ¹/₂ = -0.5

=> x = -0.5 is the axis of symmetry of the equation.

Step 2:

When x = -0.5,

f(x) = f(-0.5) = (-0.5)² + (-0.5) - 6 = 0

=> f(-0.5) = 0.25 - 0.5 - 6 = -6.25

So, the coordinates of the vertex are (-0.5, -6.25).

Step 3:

Now, make the table with all other points.

Put x = -4, -3, -2, 0, 1, 2, 3 and find f(x).

When x = -4,

f(x) = f(-4) = (-4)² + (-4) - 6 = 16 - 4 - 6 = 6

When x = -3,

f(x) = f(-3) = (-3)² + (-3) - 6 = 9 - 3 - 6 = 0

When x = -2,

f(x) = f(-2) = (-2)² + (-2) - 6 = 4 - 2 - 6 = -4

When x = -1,

f(x) = f(-1) = (-1)² + (-1) - 6 = 1 - 1 - 6 = -6

When x = 0,

f(x) = f(0) = (0)² + (0) - 6 = 0 + 0 - 6 = -6

When x = 1,

f(x) = f(1) = (1)² + (1) - 6 = 1 + 1 - 6 = -4

When x = 2,

f(x) = f(2) = (2)² + 2 - 6 = 4 + 2 - 6 = 0

When x = 3,

f(x) = f(3) = (3)² + 3 - 6 = 9 + 3 - 6 = 6

x	-4	-3	-2	-1	-0.5	0	1	2	3
f(x)	6	0	-4	-6	-6.25	-6	-4	0	6

Hence, the ordered pairs of the quadratic function are (-4, 6), (-3, 0), (-2, -4), (-1, -6), (-0.5, -6.25), (0, -6), (1, -4), (2, 0), and (3, 6).

Step 4:

Now, plot the ordered pairs in a coordinate plane and join the points.

The point where the related function f(x) = 0 is the solution to this quadratic equation. This occurs at the x-intercepts. The x-intercepts are the points where the graph touches the x-axis. So, the graph has -3 and 2 as x-intercepts.

So, the solutions are -3 and 2.

Step 5:

Check:

We can also check our solutions by factoring the given equation.

x² + x - 6 =0

=> x² + 3x - 2x - 6 = 0

=> x(x + 3) - 2(x + 3) = 0

=> (x + 3)(x - 2) = 0

Using the zero product property,

x + 3 = 0 or x - 2 = 0

x = -3 or x = 2

Step 6:

So, the solutions of the equation are -3 and 2.

Example 2:

Solve x² - 10x + 25 = 0 by graphing.

Solution:

Step 1:

Given equation is x² - 10x + 25 = 0.

Here a = 1, b = -10, and c = 25.

Graph the related function f(x) = x² - 10x + 25.

The equation of the axis of symmetry is x = - ^b/_2a

x = - ^b/_2a = - ¹⁰/₂₍₁₎ = - ^b/_2a = 5

=> x = 5 is the axis of symmetry of the given equation.

Step 2:

When x = 5,

f(x) = f(5) = (5)² - 10(5) + 25 = 25 - 50 + 25 = 0

=> f(x) = 0

So, the coordinates of the vertex are (5, 0).

Step 3:

Now, make the table with all other points.

Put x = 2, 3, 4, 6, 7, 8 and find f(x).

When x = 2,

f(x) = f(2) = (2)² - 10(2) + 25 = 4 - 20 + 25 = 9

When x = 3,

f(x) = f(3) = (3)² - 10(3) + 25 = 9 - 30 + 25 = 4

When x = 4,

f(x) = f(4) = (4)² - 10(4) + 25 = 16 - 40 + 25 = 1

When x = 6,

f(x) = f(6) = (6)² - 10(6) + 25 = 36 - 60 + 25 = 1

When x = 7,

f(x) = f(7) = (7)² - 10(7) + 25 = 49 - 70 + 25 = 4

When x = 8,

f(x) = f(8) = (8)² - 10(8) + 25 = 64 - 80 + 25 = 9

x	2	3	4	5	6	7	8
f(x)	9	4	1	0	1	4	9

So, the ordered pairs of the quadratic function are (2, 9), (3, 4), (4, 1), (5, 0), (6, 1), (7, 4), and (8, 9).

Step 4:

Now, plot the ordered pairs in a coordinate plane and join the points.

The resultant graph will be as shown.

The point where the related function f(x) = 0 is the solution to this quadratic equation. This occurs at the x-intercepts. The x-intercepts are the points where the graph touches the x-axis.

Notice that the vertex of the parabola is the x-intercept.

Hence, the solution is 5.

Step 5:

Check:

Let us check it by factoring.

Factor the equation x² - 10x + 25 = 0.

x² - 5x - 5x + 25 = 0

=> x(x - 5) - 5(x - 5) = 0

=> (x - 5)(x - 5) = 0

=> x - 5 = 0 or x - 5 = 0

=> x = 5 or x = 5

So, there are two roots but they are identical.

So, this quadratic equation has only one root called 5 as double root.

Step 6:

So, the solution is 5.

Example 3:

Solve x² + 2x + 5 = 0 by graphing.

Solution:

Step 1:

Given equation is x² + 2x + 5 = 0.

Here, a = 1, b = 2, and c = 5.

Graph the related function f(x) = x² + 2x + 5.

The equation of the axis of symmetry is x = - ^b/_2a

x = - ^b/_2a= - ²/₂₍₁₎ = -1

=> x = -1 is the axis of symmetry of the given equation.

Step 2:

When x = -1,

f(x) = f(-1) = (-1)² + 2(-1) + 5 = 1 - 2 + 5 = 4

So, the coordinates of the vertex are (-1, 4).

Now, make the table with all other points.

Put x = -4, -3, -2, 0, 1, 2 and find f(x).

When x = -4,

f(x) = f(-4) = (-4)² + 2(-4) + 5 = 16 - 8 + 5 = 13

=> f(-4) = 13

When x = -3,

f(x) = f(-3) = (-3)² + 2(-3) + 5 = 9 - 6 + 5 = 8

=> f(-3) = 8

When x = -2,

f(x) = f(-2) = (-2)² + 2(-2) + 5 = 4 - 4 +5 = 5

=> f(-2) = 5

When x = 0,

f(x) = f(0) = (0)² + 2(0) + 5 = 5

=> f(0) = 5

When x = 1,

f(x) = f(1) = (1)² + 2(1) + 5 = 1 + 2 + 5 = 8

=> f(1) = 8

When x = 2,

f(x) = f(2) = (2)² + 2(2) + 5 = 13

=> f(2) = 13

x	-4	-3	-2	-1	0	1	2
f(x)	13	8	5	4	5	8	13

So, the ordered pairs of the quadratic function are (-4, 13), (-3, 8), (-2, 5), (-1, 4), (0, 5), (1, 8), and (2, 13).

Step 3:

Now, plot the ordered pairs in a coordinate plane and join the points.

The resultant graph will be as shown.

Step 4:

So, there are no real number solutions for the given quadratic equation.

Here is some practice questions: Quadratic equations

To check your result: Quadratic equation solver

Related Articles:

>> Quadratic formula method to solve quadratic equations

>> Perfect Square Method: Detailed Review

>> Solving Quadratic Equations: Perfect Square Method

>> Solving Quadratic Equations: Factoring Method

>> Solving Inequalities: Graphing Method

>> Solving Linear Equations: Graphing Method

>> Linear Equations: Word Problems